A very small earthed conducting sphere is at a distance $a$ from a point charge $q_1$ and at a distance $b$ from a point charge $q_2(a<b)$. At a certain instant, the sphere starts expanding so that its radius grows according to the law $R=vt$. Determine the time dependence $I\left(t\right)$of the current in the earthing conductor, assuming that the point charges and the centre of the sphere are at rest, and in due time the initial point charges get into the expanding sphere without touching it (through small holes). 

Let us write the condition of the zero potential of the sphere, and hence of any point inside it (in particular, its centre), at any instant of time $t$. We shall find out three times intervals:

(1) $t<\frac{a}{v}$ , (2) $\frac{a}{v}⩽t<\frac{b}{v}$ , (3) $t⩾\frac{b}{v}$

Denoting the charge of the sphere by $q\left(t\right)$, we obtain the following expression for an instant $t$ from the first time interval:

$\frac{q_1}{a}+\frac{q_2}{b}+\frac{q\left(t\right)}{vt}=0$, 

Hence, $$\begin{gathered} q\left(t\right)=-v\left(\frac{q_1}{a}+\frac{q_2}{b}\right)t, \\ {I}_1\left(t\right)=-v\left(\frac{q_1}{a}+\frac{q_2}{b}\right) \end{gathered}$$

For an instant $t$ from the second time interval, we find that the fields inside and outside the sphere are independent, and hence 

$$\begin{gathered} \frac{q\left(t\right)+q_1}{vt}=\frac{q_2}{b}, \\ {I}_2\left(t\right)=-v\frac{q_2}{b} \end{gathered}$$

Finally, as soon as the sphere absorbs the two point charges $q_1$ and $q_2$, the current will stop flowing through the earthing conductor, and we can write $I_3\left(t\right)=0$. 

Thus, 

$I\left(t\right)=\left\{\begin{array}{l}-v\left(\frac{q_1}{a}+\frac{q_2}{b}\right), t<\frac{a}{v},\\ -v\frac{q_2}{b}, \frac{a}{v}\le t<\frac{b}{v}\\ 0, t\ge \frac{b}{v} \end{array}\right.$